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Question: Find the derivative of \[f(x) = \dfrac{1}{x}\]...

Find the derivative of f(x)=1xf(x) = \dfrac{1}{x}

Explanation

Solution

First we have to know that the derivative of f(x)f(x),
The rate at which a function changes with respect to independent variables is called the derivative of the function.
The derivative of a function is defined in terms of the limit involving increments of the independent and dependent variables.
Here we have to find the derivative of the given term by using limit terms.

Formula used: The derivative of a function f(x)f(x) at a point (x,f(x))(x,f(x)) is written as f′(x)f'(x) and is defined as a limit terms, that is f′(x)=lim⁡h→0f(x+h)−f(x)h{f'}(x) = \mathop {\lim }\limits_{h \to 0} \dfrac{{f\left( {x + h} \right) - f\left( x \right)}}{h}

Complete step-by-step answer:
It is given that the function f(x)=1x....(1)f(x) = \dfrac{1}{x}....\left( 1 \right)
To find its derivative
Now we have to use the formula, f′(x)=lim⁡h→0f(x+h)−f(x)h{f'}(x) = \mathop {\lim }\limits_{h \to 0} \dfrac{{f\left( {x + h} \right) - f\left( x \right)}}{h}
In the given data, f(x)=1xf(x) = \dfrac{1}{x}and f(x+h)=1x+hf(x + h) = \dfrac{1}{{x + h}}
Substitute the values to the formula we get,
\Rightarrow$$${f'}(x) = \mathop {\lim }\limits_{h \to 0} \dfrac{{\left( {\dfrac{1}{{x + h}}} \right) - \dfrac{1}{x}}}{h}$$ Taking LCM in the numerator terms we get, \Rightarrow{f'}(x) = \mathop {\lim }\limits_{h \to 0} \dfrac{{\dfrac{{x - (x + h)}}{{x.(x + h)}}}}{h}$$ On multiplying $\left( - \right)$ in the numerator term, we get $ \Rightarrow{f'}(x) = \mathop {\lim }\limits_{h \to 0} \dfrac{{\dfrac{{x - x - h}}{{x.(x + h)}}}}{h} On subtracting the term we get, $ \Rightarrow$$${f'}(x) = \mathop {\lim }\limits_{h \to 0} \dfrac{{\dfrac{{ - h}}{{x.(x + h)}}}}{h}
We can take the denominator terms as multiplying in the above denominator terms we get,
\Rightarrow$$${f'}(x) = \mathop {\lim }\limits_{h \to 0} \dfrac{{ - h}}{{xh(x + h)}}$$ Cancel the numerator h by denominator h \Rightarrow{f'}(x) = \mathop {\lim }\limits_{h \to 0} \dfrac{{ - 1}}{{x(x + h)}}$$ Substitute the limit, where, h approaches zero, That is, $$h = 0$$ $ \Rightarrow{f'}(x) = \dfrac{{ - 1}}{{x(x + 0)}} On adding we get, $ \Rightarrow$$$f'(x) = \dfrac{{ - 1}}{{x(x)}}
Let us multiply the terms we get,
⇒ \Rightarrowf′(x)=−1x2f'(x) = \dfrac{{ - 1}}{{{x^2}}}

Hence the derivative of f(x)=1xf(x) = \dfrac{1}{x} is f′(x)=−1x2f'(x) = \dfrac{{ - 1}}{{{x^2}}}.

Note: If you instead use yy to denote a function of a variable, y=f(x)y = f\left( x \right)
An alternative method of finding the derivative of the given function f(x)f(x)
Then you use the notation dydx\dfrac{{dy}}{{dx}} to denote the derivative.
The notation stems from the definition $$$$
dydx=ddx(f(x))\dfrac{{dy}}{{dx}} = \dfrac{d}{{dx}}(f(x))
Here f(x)=1xf(x) = \dfrac{1}{x}
\Rightarrow$$$\dfrac{{dy}}{{dx}} = \dfrac{d}{{dx}}\left( {\dfrac{1}{x}} \right)$$ Rewrite, $$\dfrac{1}{x}$$ as $${x^{ - 1}}$$ \Rightarrow\dfrac{d}{{dx}}({x^{ - 1}})$$ Differentiate using the Power Rule which states that, $$\dfrac{d}{{dx}}({x^n})$$is $$n{x^{n - 1}}$$ Here, $$n = - 1$$ $ \Rightarrow\dfrac{d}{{dx}}({x^{ - 1}}) = ( - 1){x^{ - 1 - 1}} On adding the power terms we get, $ \Rightarrow$$$\dfrac{d}{{dx}}({x^{ - 1}}) = ( - 1){x^{ - 2}}
⇒ \Rightarrow$$\dfrac{d}{{dx}}({x^{ - 1}}) = - {x^{ - 2}}Rewritetheexpressionusingthenegativeexponentrule, Rewrite the expression using the negative exponent rule,{b^{ - n}} = \dfrac{1}{{{b^n}}}Sowecanwriteitas, So we can write it as, - {x^2} = \dfrac{1}{{{x^2}}}Hencethederivativeof Hence the derivative off(x) = \dfrac{1}{x}isisf'(x) = \dfrac{{ - 1}}{{{x^2}}}$$.